MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 6

More number of oxidation states are exhibited by the actinoids than by lanthanoids. The main reason for this is:

0%
greater metallic character of the lanthanoids than that of the corresponding actinoids.
0%
more active nature of the actinoids.
0%
more energy difference $5f$ and $6d$ -orbitals than that between $4f$ and $5d$ -orbitals.
0%
lesser energy difference between $5f$ and $6d$ -orbitals than that between $4f$ and $5d$ -orbitals.

Titanium shows magnetic moment of

$1.73BM$ ; then find out its oxidation state:

0%
$+1$
0%
$+4$
0%
$+3$
0%
$+2$

Explanation

$1.73 BM$

$= \sqrt {n\times(n+2)}$ .

Solving this we get n

$=$ 1.

So Titanium has a electron configuration of [Ar]

$4s^03d^1$ . So its oxidation state is +3.

Magnetic moments of

$Cr(Z=24)$ ,

${Mn} (Z=25)$ and

${Fe}(Z=26)$ , are

$x ,y , z$ .

They are in order:

0%
$x< y< z$
0%
$x=y< z$
0%
$x>y>z$ .
0%
$x=y=z$

Explanation

n = number of unpaired electrons. Greater the number of unpaired electrons, greater will be the magnetic moment.

$Cr$ = 6 unpaired electron

$Mn$ = 5 unpaired electron

$Fe$ = 4 unpaired electron

Therefore the order of magnetic moment is

$x>y>z$ .

Hence,option C is correct.

Which of the following elements do not show the properties characteristics of

$d$ -block elements:

0%
$Cu,Ag,Au$
0%
$Zn,Hg,Cd$
0%
$Sc,Ti,V$
0%
$Fe,Co,Ni$

The compound of vanadium has magnetic moment of

$1.73B$ . The vanadium chloride has the formula:

0%
$V{Cl}_{2}$
0%
$V{Cl}_{3}$
0%
$V{Cl}_{4}$
0%
$V{Cl}_{5}$

Explanation

$\sqrt {n(n+2)} = 1.73$ Solving this we get the number of unpaired electrons as 1.

$V^{4+}$ will result in 1 unpaired electron. Hence the solution is

$VCl_4$

Lanthanoids contraction is caused due to:

0%
the same effective nuclear charge from $Ce$ to $Lu$
0%
the imperfect shielding of outer electrons by $4f$ electrons from the nuclear charge
0%
the appreciable shielding of outer electrons by $4f$ -electrons from the nuclear charge
0%
the appreciable shielding of outer electrons by $5d$ -electrons from the nuclear charge

Which of the following belongs to the actinoids series of elements?

0%
$Y$
0%
$Ta$
0%
$U$
0%
$P$

Explanation

$_{92}U$ , Uranium, is a part of the actinoids series of elements.

Maximum magnetic moment is shown by:

0%
${d}^{5}$
0%
${d}^{6}$
0%
${d}^{7}$
0%
${d}^{8}$

Explanation

Magnetic moment increases with the increase in number of unpaired electrons.

$d^5$ has 5 unpaired electrons .

$d^6$ has 4 unpaired electrons

$d^7$ has 3 unpaired electrons

$d^8$ has 2 unpaired electrons.

Hence,option A is correct.

The "Spin-only" magnetic moment [in units of Bohr magneton,

$({\mu}_{B})]$ of

${Ni}^{2+}$ in aqueous solution would be (At. no.

$Ni=28$ ):

0%
$0$
0%
$1.73$
0%
$2.84$
0%
$4.90$

Explanation

The electronic configuration of

$Ni^{+2}$ is

$1s^22s^22p^63s^23p^63d^8$

Number of unpaired electrons in

$Ni^{+2}$ is 2 formula used for the spin only magnetic moment

$\mu = \sqrt{n(n+2)}$

$\mu = \sqrt{2(2+2)}$

$\mu = \sqrt{8}$

$\mu = 2.84BM$

Hence, the correct option is

$\text{C}$

Which of the following elements involves gradual filling of

$5f$ -level?

0%
Lanthanoids
0%
Actinoids
0%
Transition metals
0%
Coinage metals

Explanation

Actinoids involves gradual filling of

$5f$ -level.

Hence option B is correct.

1487406_219426_ans_4ca7a64c938b456591a74108675bd71f.png

${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }$ every

$Cr$ is linked to:

0%
two $O$ -atoms.
0%
three $O$ -atoms.
0%
four $O$ -atoms.
0%
five $O$ -atoms.

Explanation

Every

$Cr$ is linked to four other

$O$ atoms in

$Cr_2O_7^{2-}$ .

Which one of the following ionic species will not impart color to an aqueous solution?

0%
${Ti}^{4+}$
0%
${Cu}^{+}$
0%
${Zn}^{2+}$
0%
${Cr}^{3+}$

Explanation

The transition metal ions which have completely filled d-orbitals are colorless.

The transition metal ions which have completely empty d-orbitals are also colorless.

$Cr^{3+}$ has a blue-green color. because contain unpair electrons.

Rest all are colorless.

In the dichromate dianion:

0%
four $Cr-O$ bonds are equivalent.
0%
six $Cr-O$ bonds are equivalent.
0%
all $Cr-O$ bonds are equivalent.
0%
all $Cr-O$ bonds are non-equivalent.

Explanation

Hint: If the bonds are in resonance then they will have the same bond length so they are equivalent.

Correct Option: $B$

Explanation:

In the $Dichromate \ di-anion$ , $6 \ Cr-O$ bonds are equivalent.
The $Dichromate$ ion consists of two $tetrahedral$ chromate sharing a $Oxygen$ at the corner.
The $six$ terminal bonds have the same bond length due to resonance.
The bonds which are named $1$ to $6$ are in resonance and due to which these bonds are equal.

The maximum number of identical Cr-O bond lengths in Cr_2 O_7^(2-) would be

Which of the following is ferromagnetic.

0%
Cu, Ag, Au
0%
Fe, Co, Ni
0%
Zn, Cd, Hg
0%
Ca, Sr, Ba

Explanation

Explanation:

Ferromagnetic substances are those which get strongly magnetized in the presence of a magnetic field.

These substances have domains that get aligned in the magnetic field, thus increasing their strength.

These are strongly attracted by a magnet.

Examples include $iron, cobalt, nickel, etc.$

Hence, option B is the correct answer.

The magnetism of

$Ni^{2+}$ ion is:

0%
$\cfrac {eh}{4\pi m}$
0%
$\sqrt 6 \left (\cfrac {eh}{4\pi m} \right )$
0%
$\sqrt 8 \left (\cfrac {eh}{4\pi m} \right )$
0%
$\sqrt {10} \left (\cfrac {eh}{4\pi m} \right )$

Explanation

$Ni^{2+}$ has valence shell electronic configuration of

$3d^84s^0$ . It contains 2 unpaired electrons. Its magnetic moment is given by:

$\displaystyle \sqrt {n(n+2)} \left (\cfrac {eh}{4\pi m} \right )$

$= \sqrt {2(2+2)} \left (\cfrac {eh}{4\pi m} \right )$

$= \sqrt 8 \left (\cfrac {eh}{4\pi m} \right )$

Elements with incomplete f-subshell are called:

0%
d-block elements
0%
outer transition elements
0%
inner transition elements
0%
f-block elements

The incorrect statement in respect of chromyl chloride test is?

0%
Formation of red vapour
0%
Formation of leas chromate
0%
Formation of carbonyl chloride
0%
Liberation of chlorine

Explanation

Hint: chromyl chloride test refers to a chemical test for the detection of chloride ions qualitatively.

Chromyl chloride test is a qualitative analysis test used for the conformation for

$\mathrm{Cl}^{-}$ ions.

Let us understand the reaction involved

Step 1: Take a small amount of soluble chloride salt with an equal amount of powdered

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ .

Step 2: Heat the mixture with conc.

$\mathrm{H}_{2} \mathrm{SO}_{4}$ .

$\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+4 \mathrm{KCl}+6 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+6 \mathrm{KHSO}_{4}+3 \mathrm{H}_{2} \mathrm{O}$

Step 3: The orange red vapours of formed chromyl chloride

$\left(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\right)$ is passed through the

$\mathrm{NaOH}$ solution taken in another test tube.

$\mathrm{CrO}_{2} \mathrm{Cl}_{2}+4 \mathrm{NaOH} \rightarrow 2 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{NaCl}+2 \mathrm{H}_{2} \mathrm{O}$

Step 4: The solution turns yellow.

Step 5: Add acetic acid to the solution and then lead acetate solution.

$\mathrm{Na}_{2} \mathrm{CrO}_{4}+\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{~Pb} \rightarrow \mathrm{PbCrO}_{4}+2 \mathrm{CH}_{3} \mathrm{COONa}$

Step 6: Yellow PPT

$\left(P b C r O_{4}\right)$ is formed.

So, from the above information, we can see there is no liberation of chlorine.

Final Answer:

Hence D is correct.

Out of the ions

$Ag^+, Co^{2+}, Ti^{4+}$ , which one will be coloured in aqueous solutions.

[Atomic no : Ag = 47, Co = 27, Ti = 22]

0%
$Ag^+$
0%
$Co^{2+}$
0%
$Ti^{4+}$
0%
None of these

Explanation

compound with unpaired d-electron will show color.

$Ag^+ = 5d^{10} \,\,\,\,\,\,,n = 0$

$Ti^{+4} = 3s^2 3p^6 \,\,\,\,\,\,, n = 0$

$Co^{+2} = [Ar]3d^7\,\,\,\,\,\,\,,n = 3$ It will show color due to unpaired d electrons.

One unpaired electron in atom contributes a magnetic moment of 1.1 BM. The magnetic moment of

$Cr$ (Atomic number 24) is:

0%
4.6
0%
5.6
0%
6.6
0%
7.6

Explanation

$Cr\left( z=24 \right) \Rightarrow \left[ Ne \right] { 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 1 }{ 3d }^{ 5 }$

$n=6$

Now,

$\mu =\sqrt { n\left( n+2 \right) }$

$\mu =\sqrt { 6\left( 8 \right) } =\sqrt { 48 } =6.6BM$

The ion that gives colourless compound:

0%
$Sc^{+3}$
0%
$V^{+3}$
0%
$Cr^{+3}$
0%
All of these

The electronic configuration of

$Co^{2+}$ and

$Cu^{2+}$ is

$d^7$ and

$d^9$ respectively,which of these ions is expecteed to be more paramagnetic ?

0%
$Co^{+2}$
0%
$Cu^{2+}$
0%
Both have same
0%
None of these

Explanation

As we know,

$\mu = \sqrt {n(n+2)}$
where n = no of unpaired electron.

$Co^{+2}$ = more paramagnetic

$d^7 ; n = 3$

$Cu^{+2} = d^9 ; n = 1$

Which one of the following paramagnetic in nature?

0%
$CuCl_2$
0%
$CaCl_2$
0%
$CdCl_2$
0%
All of these

Explanation

Option

$(A)$ is correct.

$CuCl_2$ ,

$Cu$ is in +2 oxidation state.

Electronic configuration of

$Cu^{2+}$ is

$[Ar], 3d^9$ , and is coloured and paramagnetic due to presence of an unpaired electron in the

$3d$ sub-orbital.

The chromate ion in acidic medium changes to ..............

0%
$CrO_4^{-2}$
0%
Cr $_3$ O $_{10}^{-2}$
0%
Cr $_2$ O $_7^{-2}$
0%
none of these

Explanation

Option (C) is correct.

The chromate ion (yellow colour) in acidic medium changes to dichromate ion

$Cr_2O_7^{2-}$ (orange - red colour). In alkaline medium, reverse happens.

In solution,the dichromate ions convert to chromate ions by the action of a base. Thus chromate ion and dichromate ion exists in equilibrium.

Which of the following ion is not coloured ?

0%
$Ni(DMG)_2$
0%
$[Co(SCN)_4]^{2-}$
0%
$[Fe(H_2O)_5SCN]^{2+}$
0%
$[Al(OH)_4]^-$

Explanation

$[Al(OH)4]^-$ is not colored because the transition of electrons is not possible in this.
Whereas, other are colored due to being transition metal complex have electrons in d orbitals and can undergo d-d transition hence can impart color.

The transition metal used in X-rays tube is :

0%
$Mo$
0%
$Ta$
0%
$Tc$
0%
$Pm$

If each one of the above ionic species is in turn kept in a magnetic field, which one will get attracted ?

0%
$Co^{+2}$
0%
$Mn^{+7}$
0%
$Sc^{+3}$
0%
$Ti^{+4}$

Explanation

$Co^{+2} = [Ar]3d^7\,\,\,\,\,\,\,n = 3$ attracted by a magnetic field due to presence of 3 unpaired electron.

Raney nickel, which is used as a catalyst, is obtained:

0%
by heating nickel tetracarbonyl
0%
by grinding solid nickel
0%
by treating nickelaluminium alloy with dilute $NaOH$
0%
by treating nickelaluminium alloy with dilute $HCl$

Explanation

Raney nickel, s solid catalyst composed of fine grains of a nickel Aluminium alloy which is used in many industrial process. It is obtained by treating nickel-Aluminium alloy with dilute

$NaOH$ .

Cholrine gas is produced from HCI by the addition of:

0%
$K_2SO_4$
0%
$KCl$
0%
$KNO_3$
0%
$KMnO_4$

Explanation

$KMnO_4$ being strong oxidising agent will oxidise CI

$^-$ ion present in HCl to form chlorine gas.

$2KMnO_4 + 16 HCl \rightarrow 2KCl+ 2MnCl_2 + 8H_2O + 5Cl_2$

Most copper

$(I)$ compounds are found to be colourless. This is due to :

0%
presence of low oxidation state of copper
0%
completely filled d-level in $Cu(I)$
0%
diamagnetic nature of the compound
0%
high polarizability of $Cu(I)$ ion

The metal capable of gaining as well as losing an electron is:

0%
Hg
0%
Ca
0%
Ag
0%
Au

In the general electronic configuration,

$(n-2)f^{1-14} (n-1)d^{01} ns^2$ (where n = 7) the configuration will be of:

0%
lanthenides
0%
actinides
0%
transition elements
0%
none

Explanation

General electronic configuration is given:

$(n -2) f^{1 -14} (n -1) d^{0 1} ns^2 ,where ( n =7 )$

$5f^{1 -14} 6d^{0 1 }ns^2$
The seventh period

$(n=7)$ is similar to the sixth period with the successive filling up of the 7s, 5f, 6d and 7p orbitals and includes most of the man-made radioactive elements. This period will end at the element with atomic number 118 which would belong to the noble gas family. Filling up of the 5f orbitals after actinium

$(Z=89)$ gives the 5f-inner transition series known as the Actinoid Series.

Which of the following is not a heavy metal?

0%
$Cu$
0%
$Hg$
0%
$Pb$
0%
$K$

Explanation

A heavy metal is any relatively dense metal. Density criteria range from above

$3.5g/cm^3$ to above

$7g/cm^3$ . Atomic numbers of heavy metals are generally greater than 20.

$Cu, Hg$ and

$Pb$ are heavy metals but

$K$ is not a heavy metal. So answer is D.

The decreasing order of solubility of silver halide is:

0%
$AgI > AgBr > AgCl > AgF$
0%
$AgF > AgCl > AgBr > AgI$
0%
$AgCl > AgF > AgBr > AgI$
0%
$AgBr > AgF > AgI > AgCl$

Explanation

The decreasing order of solubility of silver halide is

$\displaystyle AgF > AgCl > AgBr > AgI$
On moving down the group of halogens, the ionic size increases and hydration energy decreases. Hence, the solubility decreases.

The tendency towards complex formation is maximum in:

0%
s-block elements
0%
p-block elements
0%
d-block elements
0%
none of the above

Explanation

The tendency towards complex formation is maximum in

$d-block$ elements as they have maximum vacant orbitals and thus shows variable oxidation state.

Which of the following ions is colourless in solution?

0%
${V}^{3+}$
0%
${Cr}^{3+}$
0%
${Co}^{2+}$
0%
${Sc}^{3+}$

Explanation

$V\left( 23 \right) =\left[ Ar \right] 3{ d }^{ 3 },4{ s }^{ 2 }$

${ V }^{ 3+ }=\left[ Ar \right] 3{ d }^{ 2 },4{ s }^{ 0 }$ (two unpaired electrons)

$Cr\left( 24 \right) =\left[ Ar \right] 3{ d }^{ 5 },4{ s }^{ 1 }$

${ Cr }^{ 3+ }=\left[ Ar \right] 3{ d }^{ 3 },4{ s }^{ 0 }$ (three unpaired electrons)

$Co\left( 27 \right) =\left[ Ar \right] 3{ d }^{ 7 },4{ s }^{ 2 }$

${ Co }^{ 2+ }=\left[ Ar \right] 3{ d }^{ 7 },4{ s }^{ 0 }$ (three unpaired electrons)

$Sc\left( 21 \right) =\left[ Ar \right] 3{ d }^{ 7 },4{ s }^{ 2 }$

${ Sc }^{ 3+ }=\left[ Ar \right] 3{ d }^{ 0 },4{ s }^{ 0 }$ (no unpaired electrons)
Thus, in

${ Sc }^{ 3+ }$ no unpaired d-electron is present.
Hence, no d-d transition is possible and it is colourless ion.

The pair that referred as 'chemical twins' is:

0%
Ac, Cf
0%
Hf, Ta
0%
Tc, Re
0%
La, Ac

Explanation

Due to the presence of

$f$ -electrons, the nuclear is not screened effectively. Thus, the atom radius of elements of second transition series (i.e., Nb, Mo, Tc, etc) is similar to that of corresponding third transition series elements (i.e., Ta, W, Re).

Hence, Nb-Ta; Mo-W or Tc-Re are called chemical twins.

Generally transition elements and their salts are coloured due to presence of___________

0%
coloured compounds and ions
0%
coloured anions and cations
0%
coloured particles and ions
0%
unpaired electrons

Which of the following is coloured compound?

0%
$CuF_2$
0%
$CuI$
0%
$NaCl$
0%
$MgCl_2$

Explanation

The d block elements having unpaired electrons form coloured compound in

$CuF_2$ copper is

$Cu^{++}$ state.

$Cu^{++} = 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^0, 3d^9$ .

It has one unpaired electron. So,

$CuF_2$ is coloured compound.

Select a ferromagnetic material from the following?

0%
Dioxygen
0%
Chromium (IV) oxide
0%
Benzene
0%
Dihydrogen monoxide

Most basic hydroxide among the following is:

0%
$Lu(OH)_3$
0%
$Eu(OH)_3$
0%
$Yb(OH)_3$
0%
$Ce(OH)_3$

Explanation

Due to lanthanide contraction, the size of

$M^{3+}$ ions

$[i.e, Lu^{3+}, Eu^{3+}, Yb^{3+}$ and

$Ce^{3+}]$ decreases and thus, the basic strength of their hydroxides decreases. The order of size of given

$M^{3+}$ ions is

$Ce^{3+}> Eu^{3+}> Yb^{3+} > Lu^{3+}$

$\therefore$ The order of basic strength of hydroxides is

$Ce(OH)_3 > Eu(OH)_3 > Yb(OH)_3 > Lu(OH)_3$ (most basic).

Which one of the following metal ion is coloured?

0%
$Ti^{2+}$
0%
$Cu^+$
0%
$V^{4+}$
0%
$Zn^{2+}$

Explanation

The metal ion

$\displaystyle V^{4+}$ is coloured.

Atomic number of Vanadium is 23

Electronic configuration of V

$\displaystyle =[Ar]4s^23d^3$

Electronic configuration of

$\displaystyle V^{+4}=[Ar]3d^1$

It contains one unpaired electron hence it is coloured.

Hence the correct option is C.

The point of dissimilarity between lanthanides and actinides is:

0%
three outermost shells are partially filled
0%
they show oxidation of $+3$ ( common)
0%
they are called inner transition elements
0%
they are radioactive in nature

Which of the following commonly forms an ionic compounds as a

$-1$ anion?

0%
$\displaystyle C$
0%
$\displaystyle N$
0%
$\displaystyle O$
0%
$\displaystyle Fe$
0%
$\displaystyle Ne$

Explanation

${ Na }_{ 2 }O$ is an ionic compound

${ Na }^{ 2+ }\quad { O }^{ 2- }\ \rightarrow$ Oxygen is forming ionic compound as an anion.

The spin only magnetic moment of

$Ni^{2+}$ (in B.M.) in aqueous solution will be:

0%
$1.73$
0%
$2.82$
0%
$3.42$
0%
$2.25$

Explanation

$\displaystyle Ni^{2+}:\ [Ar]3d^8$ has 2 unpaired electrons.

The spin only magnetic moment,

$\displaystyle \mu=\sqrt{n(n+2)}$

$=\sqrt{2(2+2)}=\sqrt{8}$

$=2.82$ B.M

Copper (II) sulfate solution is:

0%
red
0%
purple
0%
orange
0%
green
0%
blue

Explanation

Copper (II) sulphate solution is blue in colour.
Note: Hydrated copper sulphate

$\displaystyle CuSO_4 \cdot 5 H_2O$ is blue in colour and anhydrous copper sulphate

$\displaystyle CuSO_4$ is anhydrous.

Platinum, Palladium and Iridium are called noble metals because:

0%
Alfred Nobel discovered them
0%
they are shining lustrous and pleasing to look at
0%
they are found in native state
0%
they are inert towards many common reagents

Which of the following ions has the maximum magnetic moment?

0%
$Co^{2+}$
0%
$Fe^{3+}$
0%
$Sc^{3+}$
0%
$Cr^{+3}$

Are the most readily oxidized elements within a given period

0%
3d Transition metals
0%
Alkali metals
0%
Halogens
0%
Noble gases
0%
Actinides

Which of the following is not an actinoid?

0%
$Curium (Z = 96)$
0%
$Californium (Z = 98)$
0%
$Uranium (Z = 92)$
0%
$Terbium (Z = 65)$

Explanation

Explanation:
The actinides can be identified based on the outermost orbitals. The valence electrons of the actinides enter into the 5f-orbital.

Electronic configuration of curium:- [Rn]

${5f}^{7} {6d}^{1}$

The energy of 6d orbitals is less than the 5f orbital, so the electrons first enter into 6d orbitals, and then the remaining valence electrons enter into 5f orbital. Thus, Curium is an actinoid.

Electronic configuration of californium:- [Rn]

${5f}^{10} {7s}^{2}$

The energy of 7s orbitals is less than the 5f orbital, so the electrons first enter into 7s orbitals, and then the remaining valence electrons enter into 5f orbital. Thus, Californium is an actinoid.

Electronic configuration of uranium:- [Rn]

${5f}^{3}$

The valence electrons of the uranium element enter into 5f-orbital so the uranium is an actinoid.

Electronic configuration of terbium:- [Xe]

${4f}^{9} {6s}^{2}$

The energy of 6s orbitals is less than the 4f orbital, so the electrons first enter into 6s orbitals, and then the remaining valence electrons enter into 4f orbital. Thus, Terbium is lanthanoid and not actinide.

Correct Option:

$D$

Identify the pairs of ions possessing equal spin only magnetic moments:

0%
$Co^{-2}, Ni^{+2}$
0%
$Co^{+2}, Cr^{+3}$
0%
$Fe^{+2}, Fe^{+3}$
0%
$Fe^{+2}, Cr^{+3}$

Explanation

Spin Magnetic moment is $\sqrt{n(n+2)}$

where n is the number of unpaired electrons in $Co^{+2}$ and $Cr^{+3}$ the number of unpaired electrons is $2$ . So they have same spin magnetic moment

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 6 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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